Question

Out 2000 staff 48% preferred coffee 54% tea and 64% cocoa. Of the total 28% used coffee and
tea 32% tea and cocoa and 30% coffee and cocoa. Only 6% did none of these. Find the number
having all the three.​

Answer 1

answer. 360

Explanation:Use the formula

n(AUBUC) = n(A)+n(B)+n(C) - n(AUB) - n(BUC) - n(CUA) + n(AnBnC)

//Note: The total percentage should be 100%. Also 6% did none of these. but this will need to be added so that the total percentage is 100%.

100 = 48 + 54 + 64 - 28 -32 -30 +  n(AnBnC) + 6

n(AnBnC) = 100 - 82 = 18%

So 18% had all the three.

Number of staff who had all three = 18/100 * 2000 = 360.